Let N be the set of natural numbers and R be the set of all real numbers. If function \( f:N\rightarrow R \) defined by \( f(x)=4x^{2}+12x+15 \), then \( f:N\rightarrow range(f) \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → One-one and onto
Completing the square, \( f(x)=(2x+3)^{2}+6 \).
One-one: if \( f(x_{1})=f(x_{2}) \) with \( x_{1},x_{2}\in N \), then
\( \displaystyle (2x_{1}+3)^{2}=(2x_{2}+3)^{2} \)
Both \( 2x_{1}+3 \) and \( 2x_{2}+3 \) are positive, so \( 2x_{1}+3=2x_{2}+3 \), i.e. \( x_{1}=x_{2} \).
Onto: the co-domain is now range(f), the set of all values actually taken by f, so every element of it is the image of some natural number.
Hence \( f:N\rightarrow range(f) \) is one-one and onto.
Tip: Replacing the co-domain by the range always makes a function onto; only one-one needs checking.